We are given the mass % of each.

Recall the **mass percent formula**:

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{mass}}{\mathbf{=}}\frac{\mathbf{mass}\mathbf{}\mathbf{component}}{\mathbf{total}\mathbf{}\mathbf{mass}}{\mathbf{\times}}{\mathbf{100}}}$

$\overline{){\mathbf{mass}}{\mathbf{}}{\mathbf{component}}{\mathbf{=}}\frac{\mathbf{mass}\mathbf{\%}}{\mathbf{100}}{\mathbf{\times}}{\mathbf{mass}}{\mathbf{}}{\mathbf{of}}{\mathbf{}}{\mathbf{sample}}}$

**percent fluorine: 37.42%**

Copper (II) fluoride contains 37.42% F by mass.

Use this percentage to calculate the mass of fluorine (in g) contained in 55.5 g of copper (II) fluoride.

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